Let $f(x) = 6x^{2}+8x-4$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )?
Solution: The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $6x^{2}+8x-4 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = 6, b = 8, c = -4$ $ x = \dfrac{-8 \pm \sqrt{8^{2} - 4 \cdot 6 \cdot -4}}{2 \cdot 6}$ $ x = \dfrac{-8 \pm \sqrt{160}}{12}$ $ x = \dfrac{-8 \pm 4\sqrt{10}}{12}$ $x =\dfrac{-2 \pm \sqrt{10}}{3}$